Question

Answer question plz with explanation.​

Answer 1

Answer:

Given that,

$tan\theta = 3 \\ and \\ \theta \in \: 3rd \: quad.$

GENERAL FORMULAS,

${sec}^{2}\theta-1={tan}^{2} \theta \\ \\ {sin}^{2}\theta=1-{cos}^{2}\theta$

SOLUTION:-)

As

$tan {}^{2} \theta = 9 \\ \\ {sec}^{2} \theta - 1 = 9 \\ \\ {sec}^{2} \theta = 10 \\ \\ {cos}^{2} \theta = \frac{1}{10} \\as \\ \: \: {sin}^{2} \theta = 1 - {cos}^{2} \theta \\ \\ \: \: {sin }^{2} \theta = 1 - \frac{1}{10} \\ \\ \: \: {sin}^{2} \theta = \frac{9}{10} \\ \\ sin\theta = \pm \sqrt{ \frac{9}{10} } = \\ \\ sin \theta = \pm \frac{3}{ \sqrt{10} }$

As,

θ lies in third quadrant and in third quadrant trigonometric ratio sine is negative.

So,

$sin \theta = - \dfrac{3}{ \sqrt{10} }$

$\large{\red{ So,Option\: C\: is \:correct }}$